Question

What will be the pressure exerted by a mixture of $3.2g$ of methane and $4.4g$ of carbon dioxide contained in a $9{dm}^3$ flask at ${27}^{o}C$?

• A. $8.314\times {10}^{-4}Pa$
• B. $8.314\times {10}^{-6}Pa$
• C. $6.314\times {10}^{2}Pa$
• D. $7.314\times {10}^{4}Pa$

Answer 1

Answer: (A)

$8.314\times {10}^{-4}Pa$

Solution:

Moles of $CH₄,$
$nCH₄$= Mass of CH₄/Molar mass of CH₄ [ Molar mass of CH₄] = $12+4\times 1=16$
=$3.2/16=0.2mol$
Moles of CO₂
$nCO₂$ = $4.4/44=0.1mol[MolarmassofCO₂=12+2\times 16=44]$
Total moles =$0.2+0.1=0.3mol$
$pV=nRT$
Pressure,$P=nRT/V$
$0.3mol\times 0.0821dm³atm${ k }^{ -1 }$mo${ l }^{ -1 }$\times 300K/9dm³=0.0821atm$
In terms of SI unit
Pressure,$p$ = 0.3 mol × 8.314 Pa m³$k^{-1}$ mo$l^{-1}$ × 300/ 9×10⁻³ m³
Therefore, $P=8.314$X ${10}^{-4}$ Pa is the answer.